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Bash regex for same sender and receiver with backreference

I try to make a regex (important that ist a regex because i need it for fail2ban) to match when the receiver and the sender are the same person:

echo "from=<test@test.ch>    to=<test@test.ch>" | grep -E -o  '([^=]*s)[ ]*1'

What am I doing wrong ?

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Answer

You might use a pattern to match the format of the string between the brackets with a backreference to that capture.

from(=<[^s@<>]+@[^s@<>]+>)s*to1

Explanation

  • from Match literally
  • ( Capture group 1
    • =< Match literally
    • [^s@<>]+ Match 1+ times any char except a whitespace char or @ < >
    • @ Match literally
    • [^s@<>]+ Again match 1+ times any char except a whitespace char or @ < >
    • > Match literally
  • ) Close group 1
  • s*to1 Match 0+ whitespace chars, to and the backreference to group 1

Regex demo | Bash demo

Use grep -P instead of -E for Perl compatible regular expressions.

For example

echo "from=<test@test.ch>    to=<test@test.ch>" | grep -oP  'from(=<[^s@<>]+@[^s@<>]+>)s*to1'

A bit broader match could be capturing what is between the brackets

[^=s]+(=<[^<>]+>)s*[^=s]+1

Regex demo

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